Acoustic lens patent 3,957,134 expired a long time ago. At the time it would have been prohibitively expensive to make, but now it would be reasonable with 3D printing. The primary use for it would be to disperse the output of straight axis exponential horns. In spite of the flare of the horn mouth, the wave emerging is close to a plane wave.
In the picture, plane waves enter the left side of the lens and curved waves emerge from the right side of the lens.
To my knowledge, this lens was never made and tested. It would be less expensive to make and test a small prototype. This would require a small horn tweeter. The JBL company's model 075 is no longer in production, but can be found used. It is for very high frequencies only, and must be protected from frequencies below 7000Hz by the JBL 3105 crossover network. Another alternative would be to use a horn driver and associated crossover that are both designed for frequencies above 500Hz. A straight axis circular cross section exponential horn could be 3D printed to be used with a prototype lens, but this would be a much larger lens than would work with the 075.
The principal of the lens is to delay outer parts of the emergent wave relative to the center of the wave to bend the wave. The shortest distance between two points is a straight line. The tube for the center of the wave is straight. The outer tubes are curved by varying amounts to increase delay relative to the center.
If the lens has circular symmetry the center tube would have circular cross section. The outer tubes would be concentric rings of tubes with isosceles trapezoid cross section for the length of each tube. But two of the opposite sides would be arcs, not straight lines.
The example shown has the exit ends of most of the tubes extended beyond the exit side of the lens. This is intended to reduce reflection. The exit wave is semicircular. But application requirements might dictate other choices.
This article presents a particular way to design curves with a particular amount of delay in excess of the delay of a straight tube.
The curves have symmetry about the center of each curve. The curved parts of the outer curves end at the entrance and exit planes of the lens. The ends of the other curves do not, in order to keep the curves separated. So they could be thought to end at imaginary reference planes.
The design process will be for only the exit half of a curve. The design will be with reference to a middle reference plane and an exit reference plane. The middle plane will be the same for all curves. The exit plane will be different for different curves, and will be chosen for convenience.
In the figure, point 1 is at the middle plane, as are points 3 and 4. Point 2 will be where the curved portion of the tube to be designed will end. A straight portion could extend beyond point 2.
Assuming your design calls for delay d each half of the lens. This d would be the difference between the length of line 1,2 and the length of the curve 8,2. The arc 8,2 is easy to design for a given length, but the sharp angle at 2 would create unnecessary reflection. the compound curve 8,2 would avoid the reflection, but would be harder to design. But the way the compound curve is constructed, it has exactly the same length as the arc.
First to design the arc. The arc has radius r and subtends and angle a. Use the computer programming symbol for multiplication, "*". The length is r*a, provided the angle a is expressed in radians. One radian is 57.295 degrees. First guess a position for point 7. The distance from point 7 to point 2 will be r, the radius of the arc. The angle between line 7,2 and line 7,3 will be a, the angle subtended by the arc. Calculate r*a. It will not be the value that you wanted. A computer program could rapidly improve guesses for the location of point 7 to get a satisfactory value for r*a. Then position point 8 the distance r abover point 7.
We can draw the arc 8,2.
Now that we have a simple arc with the desired length, we will design the smooth compound curve. We establish point 9 an arbitrary distance y above point 7. We establish point 10 the same distance y above point 2. Then we establish point 11 a distance y from point 10 along the line 10,9. The 11 is difficult to see in the diagram because it overlaps a line.
Point 9 will be the center of a new arc 8,11. Point 10 will be the center of a new arc 11,2.
The radius of arc 8,11 is r-y. The radius of arc 11,2 is y. Since line 9,10 is parallel to line 7,2, the angle subtended by both new arc is a, the same as the original arc. (r-y)*a + y*a = r*a. The length of the compound curve is exactly equal to the length of the simple arc.
Presumably the compound curve would be used as the center of the tube. The boundaries of the tube would be plotted relative to the center.
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